Engineering Mechanics — Rotational Inertia of a Thin Disc For a thin, solid circular disc of mass m and radius r, what is the mass moment of inertia about the axis through its center of gravity and perpendicular to the plane of the disc (the polar/shaft axis)?

Introduction / Context: In dynamics of rigid bodies, the mass moment of inertia measures resistance to angular acceleration about a chosen axis. For common shapes like discs and cylinders, standard closed-form results exist and are frequently used in design, vibration, and machine dynamics.

Given Data / Assumptions:

• Body: thin, solid disc of uniform density.
• Mass m and radius r are known constants.
• Axis: through the disc’s center of gravity and perpendicular to its plane (polar axis).

Concept / Approach: The polar mass moment of inertia I about the disc’s central axis is obtained by integrating r^2 over all elemental masses. For a thin disc, mass is distributed in the plane; symmetry simplifies the integral to a well-known formula.

Step-by-Step Solution:

Verification / Alternative check: Compare with a thin ring of radius r: I_ring = m r^2. Since a solid disc has more mass near the center, its I must be smaller; I = (1/2) m r^2 is consistent with this physical intuition.

Why Other Options Are Wrong: The factors 1/4, 1/6, and 1/8 are not produced by the correct integration for a uniform disc and would under-predict rotational inertia.

Common Pitfalls: Confusing the polar axis with a diametral axis. About a diametral centroidal axis, the inertia differs: I = (1/4) m r^2 + (1/12) m t^2 for a finite-thickness cylinder (t = length).

Final Answer: mr^2/2.